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I think differentiation in two or more variables mirrors differentiation in one variable pretty closely, swapping $\frac{d}{dx}$ operators with $\frac{\partial}{\partial x}$ usually being the only difference in process to things like the chain rule, but this idea gives me issues considering implicit differentiation.

Consider the function below:

$$2xz^3 -3yz^2+x^2y^2+4z=0$$

Where $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ must be found and $z$ is given implicitly.

My initial thought process would be to apply the $\frac{\partial z}{\partial x}$ operator on both sides of the equation and see where that brings me:

$$\frac{\partial }{\partial x}[2xz^3 -3yz^2+x^2y^2+4z]= \frac{\partial }{\partial x}[0]$$

Now, the root of my question is how to find $\frac{\partial }{\partial x}[2xz^3]$ is done where $z$ is defined implicitly. My hunch is to note that since $z$ is defined implicitly, it may or may not be a function of $x$ and thus cannot be merely treated as a constant. So, I'd say it is this due to this being a product rule:

$$\frac{\partial }{\partial x}[2xz^3] = 6z^2x\frac{\partial z}{\partial x}+z^3$$ If this turns out to be how this is done, I can solve the rest by grouping up the partial $z$ partial $x$ terms and solving for it, but I need to know if my hunch is correct, and if not, why.

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  • $\begingroup$ Yes, this is correct. (By the way you seem to have lost a factor of $2$ in your last line) $\endgroup$ – John Doe Oct 12 '17 at 22:52
  • $\begingroup$ Bleh, right. And thank you! $\endgroup$ – sangstar Oct 12 '17 at 22:59
  • $\begingroup$ Just a quibble: you’re applying the ${\partial\over\partial x}$ operator, not the ${\partial z\over\partial x}$ operator. $\endgroup$ – amd Oct 12 '17 at 23:06
  • $\begingroup$ If in doubt, write the differentiation out completely: ${\partial\over\partial x}[2xz^3]=6z^2x{\partial z\over\partial x}+2z^3{\partial x\over\partial x}$. ${\partial x\over\partial x}=1$, but ${\partial z\over\partial x}$ is as yet unknown, so it remains. $\endgroup$ – amd Oct 12 '17 at 23:08
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You can use the formulas: $$z_x=-\frac{F_x}{F_z}=-\frac{2z^3+2xy^2}{6xz^2-6yz+4};$$ $$z_y=-\frac{F_y}{F_z}=-\frac{-3z^2+2x^2y}{6xz^2-6yz+4}.$$

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